1. If L[f(t)] =	2(s + 2)	, then ƒ(0+) and ƒ(∞) are given by
   	s² + 2s + 5

1. 1. 0, 2 respectively

   
   
   

   2. 2, 0 respectively

   
   
   

   3. 0, 1 respectively

   
   
   

   4. 2/5, 0 respectively

   
   
   

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1. View Hint View Answer Discuss in Forum

   2(s + 1)	=	2(s + 1)	= 2		s + 1
   (s² + 2s + 5)		(s² + 1)² + 5			(s + 1)² + 4

   
   This is standard Laplace Transform for the function
   

   L| exp (– at) cos ωt | =	(s + a)
   	(s + a)² + ω²

   
   where a = 1, ω = 2
   Hence ƒ(t) = 2e^{– t} cos 2t
   ƒ(0) = ƒ(0+) = 2e^{– 0} cos 2 × 0 = 2
   ƒ(∞) = 2e^{– ∞} t = 0

   Correct Option: B

   2(s + 1)	=	2(s + 1)	= 2		s + 1
   (s² + 2s + 5)		(s² + 1)² + 5			(s + 1)² + 4

   
   This is standard Laplace Transform for the function
   

   L| exp (– at) cos ωt | =	(s + a)
   	(s + a)² + ω²

   
   where a = 1, ω = 2
   Hence ƒ(t) = 2e^{– t} cos 2t
   ƒ(0) = ƒ(0+) = 2e^{– 0} cos 2 × 0 = 2
   ƒ(∞) = 2e^{– ∞} t = 0

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2. The Laplace transform of a continuous-time signal x(t) is,
   

   X(s) =	5 - s
   	s² - s - 2

   
   If the Fourier transform of this signal exists, then x(t) is

1. 1. e^2t u(t) – 2e^{– t} u(t)

   
   
   

   2. – e^2t u(– t) + 2e^{– t} u(t)

   
   
   

   3. – e^2t u(– t) – 2e^{– t}u(t)

   
   
   

   4. none of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   X(s) =	5 - s	=	A	+	B
   	(s + 1)(s - 2)		(s + 1)		(s - 2)

   
   

   A =	5 - s			= - 2; and B	5 - s			= 1
   	s - 1		s=-1		s + 1		s=2

   
   

   then, X(t) = L– 1 X(s) = L– 1		- 2	+	1
   		s + 1		s - 2

   
   = [e^2t – 2e^{– t}]u(t)

   Correct Option: A

   X(s) =	5 - s	=	A	+	B
   	(s + 1)(s - 2)		(s + 1)		(s - 2)

   
   

   A =	5 - s			= - 2; and B	5 - s			= 1
   	s - 1		s=-1		s + 1		s=2

   
   

   then, X(t) = L– 1 X(s) = L– 1		- 2	+	1
   		s + 1		s - 2

   
   = [e^2t – 2e^{– t}]u(t)

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3. The closed-loop transfer function of a control system is given by
   

   C(s)	=	2(s - 1)
   R(s)		(s + 2)(s + 1)

   
   For a unit step input the output is

1. 1. – 3e^{– 2t + 4e – t – 1}

   
   
   

   2. – 3e^{– 2t} – 4e^{– t} + 1

   
   
   

   3. zero

   
   
   

   4. infinity

   
   
   

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1. View Hint View Answer Discuss in Forum

   C(s)	=	2(s - 1)	; R(s) =	1
   R(s)		(s + 2)(s + 1)		s

   
   

   ∴ C(s) =	2(s - 1)
   	(s + 1)(s + 2)

   
   Expanding in partial fractions
   

   =	k1	+	k2	+	k3
   	s		s + 1		s + 3

   
   

   ∴ k1 =	- 2	= - 1
   	2

   
   

   k2 =	2(s - 1)	|s=-1 =	2(- 2)	= 4
   	s(s + 2)		- 1(1)

   
   

   k3 =	2(s - 1)	|s=-2 =	2(- 3)	= - 3
   	s(s + 2)		- 2( - 2)

   
   

   Hence C(s) =	1	+	4	-	3
   	s		s + 1		s + 2

   
   and output, c(t) = [– 1 + 4e^{– t} – 3e^{– 2t}] u(t).

   Correct Option: A

   C(s)	=	2(s - 1)	; R(s) =	1
   R(s)		(s + 2)(s + 1)		s

   
   

   ∴ C(s) =	2(s - 1)
   	(s + 1)(s + 2)

   
   Expanding in partial fractions
   

   =	k1	+	k2	+	k3
   	s		s + 1		s + 3

   
   

   ∴ k1 =	- 2	= - 1
   	2

   
   

   k2 =	2(s - 1)	|s=-1 =	2(- 2)	= 4
   	s(s + 2)		- 1(1)

   
   

   k3 =	2(s - 1)	|s=-2 =	2(- 3)	= - 3
   	s(s + 2)		- 2( - 2)

   
   

   Hence C(s) =	1	+	4	-	3
   	s		s + 1		s + 2

   
   and output, c(t) = [– 1 + 4e^{– t} – 3e^{– 2t}] u(t).

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4. The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e^{– 2t}u(t). The response of this network to a unit step function will be

1. 1. 2[1 – e^{– 2t}]u(t)

   
   
   

   2. 4[e^{– t} – e^{– 2t}]u(t)

   
   
   

   3. sin 2t

   
   
   

   4. (1 – 4e^{– 4t})u(t)

   
   
   

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1. View Hint View Answer Discuss in Forum

   hs(t) = 4e– 2t u(t); H(s) =	4
   	s + 2

   
   For unit step, Y(s) = H(s) X(s)
   

   Y(s) =	4
   	s(s + 2)

   
   

   =	2	=	2	......since [X(s) = 1/s]
   	5		s + 2

   
   y(t) = 2[1 – e^{– 2t}] u(t)

   Correct Option: A

   hs(t) = 4e– 2t u(t); H(s) =	4
   	s + 2

   
   For unit step, Y(s) = H(s) X(s)
   

   Y(s) =	4
   	s(s + 2)

   
   

   =	2	=	2	......since [X(s) = 1/s]
   	5		s + 2

   
   y(t) = 2[1 – e^{– 2t}] u(t)

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5. Laplace transform of the signal x(t) = tu(t)^* cos 2πt u(t) will be

1. 1. 1

      s(s² + 4π²)

   
   
   

   2. 2π

      s²(s² + 4π²)

   
   
   

   3. 1

      s²(s² + 4π²)

   
   
   

   4. s³

      s²(s² + 4π²)

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   ⇒ X(s) =	1
   	s²(s² + 4π²)

   Correct Option: A

   
   

   ⇒ X(s) =	1
   	s²(s² + 4π²)

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